Let \(f(x)=x-\dfrac{1}{x}\).

**(a)** If \(f(x)+f(y)=0\), then prove or disprove that \(f(x^n)+f(y^n)=0\) for all integers \(n\).

**(b)** If \(f(x)+f(y)+f(z)=0\), then prove or disprove that \(f(x^n)+f(y^n)+f(z^n)=0\) for all integers \(n\).

**(c)** Do the above two problems with the added restriction that \(x,y,z\) are positive.

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TopNewest(a)Counterexample: \(y=-x\) and \(n=2a\) for some positive integer \(a\). \(f(x)+f(-x)=x-\dfrac{1}{x}-x+\dfrac{1}{x}=0\), but \(f(x^n)+f((-x)^n)=f(x^{2a})+f(x^{2a})=x^{2a}-\dfrac{1}{x^{2a}}+x^{2a}-\dfrac{1}{x^{2a}}\neq0\).(b)Again, the counterexample is \(y=-x\), \(z=1\), and \(n=2a\) for some positive integer \(a\). \(f(x)+f(-x)+f(1)=x-\frac{1}{x}-x+\frac{1}{x}+1-1=0\), but \(f(x^n)+f((-x)^n)+f(1^n)=f(x^{2a})+f(x^{2a})=f(1)=x^{2a}-\dfrac{1}{x^{2a}}+x^{2a}-\dfrac{1}{x^{2a}}+1-1\neq0\).(c)(part 1): Since \(x,y\) are positive and \(f(x)+f(y)=0\), \(x-\dfrac{1}{x}+y-\dfrac{1}{y}=0\), and \(x+y=\dfrac{1}{x}+\dfrac{1}{y}\). Adding the fractions on the RHS gives \(x+y=\dfrac{x+y}{xy}\) so \(1=\dfrac{1}{xy}\) or \(xy=1\). Therefore, \(f(x^n)+f(y^n)=x^n-\dfrac{1}{x^n}+y^n-\dfrac{1}{y^n}=x^n-\dfrac{1}{x^n}+\dfrac{1}{x^n}-x^n=0\). – Jeffery Li · 2 years, 9 months agoLog in to reply

EDIT: thanks, fixed! – Daniel Liu · 2 years, 9 months ago

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